Difference between revisions of "2023 AMC 8 Problems/Problem 12"
Grapecoder (talk | contribs) |
|||
| Line 1: | Line 1: | ||
| + | ==Problem== | ||
| + | The figure below shows a large white circle with a number of smaller white and shaded circles in its | ||
| + | interior. What fraction of the interior of the large white circle is shaded? | ||
| + | |||
| + | [[Image:2023 AMC 8-12|thumb|center|300px]] | ||
| + | |||
| + | <math>\textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{11}{36} \qquad \textbf{(C)}\ \frac{1}{3} \qquad \textbf{(D)}\ \frac{19}{36} \qquad \textbf{(E)}\ \frac{5}{9}</math> | ||
| + | |||
| + | ==Solution 1== | ||
First the total area of the <math>3</math> radius circle is simply just <math>9* \pi</math>. Using our area of a circle formula. | First the total area of the <math>3</math> radius circle is simply just <math>9* \pi</math>. Using our area of a circle formula. | ||
| Line 6: | Line 15: | ||
~apex304 | ~apex304 | ||
| + | |||
| + | ==Solution 2== | ||
Pretend each circle is a square. The second largest circle is a square with area <math>16\text{units}^2</math> and there are two squares in that square that each have area <math>4\text{units}^2</math> which add up to 8. Subtracting the medium-sized squares' areas from the second-largest square's area, we have <math>8\text{units}^2</math>. The largest circle becomes a square that has area <math>36\text{units}^2</math>, and the three smallest circles become three squares with area <math>8\text{units}^2</math> and add up to <math>3^2</math>. Adding the areas of the shaded regions we get <math>11</math>, so our answer is <math>\boxed{\text{(B)}\dfrac{11}{36}}</math>. | Pretend each circle is a square. The second largest circle is a square with area <math>16\text{units}^2</math> and there are two squares in that square that each have area <math>4\text{units}^2</math> which add up to 8. Subtracting the medium-sized squares' areas from the second-largest square's area, we have <math>8\text{units}^2</math>. The largest circle becomes a square that has area <math>36\text{units}^2</math>, and the three smallest circles become three squares with area <math>8\text{units}^2</math> and add up to <math>3^2</math>. Adding the areas of the shaded regions we get <math>11</math>, so our answer is <math>\boxed{\text{(B)}\dfrac{11}{36}}</math>. | ||
| Line 12: | Line 23: | ||
| − | == | + | ==Video Solution (Animated)== |
https://youtu.be/5RRo6pQqaUI | https://youtu.be/5RRo6pQqaUI | ||
~Star League (https://starleague.us) | ~Star League (https://starleague.us) | ||
Revision as of 01:42, 25 January 2023
Problem
The figure below shows a large white circle with a number of smaller white and shaded circles in its interior. What fraction of the interior of the large white circle is shaded?
Solution 1
First the total area of the 
radius circle is simply just 
. Using our area of a circle formula.
Now from here we have to find our shaded area. This can be done by adding the areas of the 
radius circles and add then take the area of the 
radius circle and subtracting that from the area of the , 1 radius circles to get our resulting complex area shape. Adding these up we will get 
Our answer is 
~apex304
Solution 2
Pretend each circle is a square. The second largest circle is a square with area 
and there are two squares in that square that each have area 
which add up to 8. Subtracting the medium-sized squares' areas from the second-largest square's area, we have 
. The largest circle becomes a square that has area 
, and the three smallest circles become three squares with area and add up to 
. Adding the areas of the shaded regions we get 
, so our answer is 
.
-claregu LaTeX edits -apex304
Video Solution (Animated)
~Star League (https://starleague.us)
